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Homemonodromy theorem

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# monodromy theorem

Let $C(t)$ be a one-parameter family of smooth paths in the complex plane with common endpoints $z_{0}$ and $z_{1}$. (For definiteness, one may suppose that the parameter $t$ takes values in the interval $[0,1]$.) Suppose that an analytic function $f$ is defined in a neighborhood of $z_{0}$ and that it is possible to analytically continue $f$ along every path in the family. Then the result of analytic continuation does not depend on the choice of path.

Note that it is *crucial* that it be possible to continue $f$ along all paths of the family. As the following example shows, the result will no longer hold if it is impossible to analytically continue $f$ along even a single path. Let the family of paths be the set of circular arcs (for the present purpose, the straight line is to be considered as a degenerate case of a circular arc) with endpoints $+1$ and $-1$ and let $f(z)=\sqrt{z}$. It is possible to analytically continue $f$ along every arc in the family except the line segment passing through $0$. The conclusion of the theorem does not hold in this case because continuing along arcs which lie above $0$ leads to $f(z_{1})=+i$ whilst continuing along arcs which lie below $0$ leads to $f(z_{1})=-i$.

## Mathematics Subject Classification

30F99*no label found*

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